HELP PLZ!! differentiating logs (1 Viewer)

[felixcthecat]

plz help.. i forgot how to differentiate the following log:

logx3 (log3 base x)

is it change it bac to 'ln' first? sumone plz tell me how to do it =)

 

[acmilan]

log₃x = lnx/ln3

 

[felixcthecat]

hmm.. then u differentiate?

lnx * 1/ln3
=1/x *3
= 3/x

is that right? i'm soo lost in differentiating logs... =.=

 

[acmilan]

Treat the ln3 as a constant, since it is.

d/dx lnx/ln3 = 1/(xln3)

 

[felixcthecat]

log₃x = lnx/ln3

waitwait.. it's log_x3... the opposite of wut u wrote above.. so should be ln3/lnx but then.... wut u do next?

the answer is : (-log3) / (x(logx)2) (the 2 is a square)


noteedit).. the log in the answer above is log base e

 

[acmilan]

Oh right sorry, read it wrong


You can use quotient rule:

d/dx ln3/lnx = (-ln3/x)/(lnx)² = -ln3/x(lnx)²

Or you can use chain rule:

d/dx ln3/lnx = d/dx ln3(lnx)^-1 = -ln3(lnx)²/x = -ln3/x(lnx)²

 

[felixcthecat]

im still thinking ><''

is ln3 still a constant?
ahh, it's so hard

 

[acmilan]

yep. If i helps, replace ln3 by a letter like a.

 

[felixcthecat]

ohhhhhhhhhhhhh, i get it now, cuz differentiating ln3 = 0.... hence it's only the 2nd half of the top left.. hehe, i get it now!!

thanks acmilan!!

btw.. how did u remember this if u did ur HSC in 2004? =Z

 

[acmilan]

Differentiation still plays a role in uni

 

[felixcthecat]

hehe. icic.. bac to studying~ thanx ^^